Question: $f(x) = \dfrac{ 1 }{ \sqrt{ 8 - \lvert x \rvert } }$ What is the domain of the real-valued function $f(x)$ ?
Answer: First, we need to consider that $f(x)$ is undefined anywhere where the radicand (the expression under the radical) is less than zero. So we know that $8 - \lvert x \rvert \geq 0$ This means $\lvert x \rvert \leq 8$ , which means $-8 \leq x \leq 8$ Next, we need to consider that $f(x)$ is also undefined anywhere where the denominator is zero. So we know that $\sqrt{ 8 - \lvert x \rvert } \neq 0$ , so $\lvert x \rvert \neq 8$ This means that $x \neq 8$ and $x \neq -8$ So we have four restrictions: $x \geq -8$ $x \leq 8$ $x \neq -8$ , and $x \neq 8$ Combining these four, we know that $x > -8$ and $x < 8$ ; alternatively, that $-8 < x < 8$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid -8< x <8\, \}$.